我想通过ajax请求返回记录
C#代码如下：
[HttpPost]
public Workoutsviewmodel GetSavedWorkoutTemplate(int WorkoutId)
{
return _db.Workouts.Where(w => w.Id == WorkoutId).Select(w => new Workoutsviewmodel
{
Tag = w.Tags.FirstOrDefault(),Name = w.Name,MaxEffort = w.MaxEffort,Id = w.Id,Description = w.Description,Compare = w.Compare,Category = w.Category
}).FirstOrDefault();
}
jQuery代码是：
function GetSavedWorkoutTemplate(obj) {
$("#Workout").hide();
var WorkoutId = $(obj).val();
if (WorkoutId == "") {
return;
}
var dto = { "WorkoutId": WorkoutId };
$.ajax({
type: "post",url: "/MemberWorkout/GetSavedWorkoutTemplate",contenttype: "application/json; charset=utf-8",data: dto,success: function (result) {
debugger;
$("#Workout").show();
},error: function (result) {
alert("Unable to update status.");
}
});
}
HTML：
@Html.DropDownList("ddlWorkout",new SelectList(ViewBag.ddlWorkout,"Id","Name"),"--Select Workout--",new { required = true,onchange = "GetSavedWorkoutTemplate(this);",style = "width:310px" })
在成功函数中我得到的结果是“TheGYM.Web.Models.Workoutsviewmodel”作为字符串而不是模型,请纠正我的错误,我想要返回一个模型.
最佳答案
在C#中,您需要使用Json()来正确地序列化对象.试试这个：
public ActionResult GetSavedWorkoutTemplate(int WorkoutId)
{
var workout = _db.Workouts.Where(w => w.Id == WorkoutId).Select(w => new Workoutsviewmodel
{
Tag = w.Tags.FirstOrDefault(),Category = w.Category
}).FirstOrDefault());
return Json(workout);
}
